Answer
(d) $100$
Work Step by Step
Intensity level of sound is increased from 20 dB to 40 dB.
Intensity level $10log\frac{I}{I_{0}}$
$I_{0}=10^{-12}W/m^{2}$
I=sound intensity.
For 20 dB,
$20=10log\frac{I_{20}}{I_{0}}$
$I_{20}=100I_{0}$
For 40 dB,
$40=10log\frac{I_{40}}{I_{0}}$
$I_{40}=10000I_{0}$
Therefore. $\frac{I_{40}}{I_{20}}=100$